If x^100 has 31 digit,then x^1000 will have how many digits?

• 10^1 has 2 digits = 10
  10^2 has 3 digits = 100
  10^3 has 4 digits = 1000
  similarly 10^y-1 has y digits
  if we have y digits in x^100 that means
  10^(y-1)=x^100
  10^(31-1)=x^100
  taking log both side
  30 log 10 = 100 log x
  30 = 100 log x
  now,
  10^(y-1)= x^1000
  y-1=1000 log x
  y-1=10*100 log x
  put value of 100 log x in this equation
  we get y-1=10*30
  y=301
  Ans=301
• 11 years agoHelpfull: Yes(96) No(3)
• 31-1=100 log x solve this, we wil get x=2
  so
  2^1000 has (1000 log 2)+1=302
  ANS: 302
  
• 11 years agoHelpfull: Yes(13) No(6)
• 302 digits
• 11 years agoHelpfull: Yes(7) No(9)
• 10^1=(1+1)=2 digit
  10^2=(2+1)=3 digit
  .
  .
  .
  10^30=(30+1)=31 digits
  therefore, 10^30=x^100
  30=100 log x
  multiplaying both side by 10
  30*10=10*100 log x
  300= 1000 log x = 300+1=301
• 10 years agoHelpfull: Yes(4) No(1)
• @pavan pls explain
  
• 11 years agoHelpfull: Yes(3) No(0)
• The same question is repeated on 21st jun delhi center....and ans is 301.
• 9 years agoHelpfull: Yes(2) No(1)
• i think 8 is wrong .......
  answer is 12 ...
  (team) (matches played) (win) (lose)
  1 14 11 3
  2 14 11 3
  3 14 11 3
  4 14 11 3
  5 14 11 3
  6 14 1 13
  7 14 0 14
  8 14 0 14
• 10 years agoHelpfull: Yes(1) No(0)
• taking log so 100log(x)=30.abcd...(a belong to {0-9})
  so log(x)=.30abcd...
  so 1000 log(x)=30a.bcd
  so no of digits [30a.bcd] +1 i.e 30a+1 so it can be 301 to 310
  
  
• 9 years agoHelpfull: Yes(1) No(0)
• @ Pavan can you please explain the answer?
• 11 years agoHelpfull: Yes(0) No(0)
• please explain the soln...
• 11 years agoHelpfull: Yes(0) No(0)
• 302
  solve using log
• 9 years agoHelpfull: Yes(0) No(0)